Coordinate Geometry Test

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Coordinate Geometry Test - 25

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Question 1
1 / -0

If the line segment joining $$\displaystyle (2, 3)$$ and $$\displaystyle(-1, 2)$$ is divided internally in the ratio $$3 : 4$$ by the line $$\displaystyle x+ 2 y=k,$$ then $$k$$ is :

A
$$\displaystyle\frac{41}{7}$$

B
$$\displaystyle\frac{5}{7}$$

C
$$\displaystyle\frac{36}{7}$$

D
$$\displaystyle\frac{31}{7}$$

Solution

If the line segment joining $$(x_{1},y_{1})$$ and $$(x_{2},y_{2})$$ is divide by internally in the ratio $$m_{1}:m_{2}$$ then the intersection point is
$$x=\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$$
and
$$y=\dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$$
so
$$x=\dfrac{3\times(-1)+4\times2}{3+4}=\dfrac{5}{7}$$
$$y=\dfrac{3\times(2)+4\times3}{3+4}=\dfrac{18}{7}$$
this point $$\left(\dfrac{5}{7},\dfrac{18}{7}\right)$$ lies on given equation of line.
satisfying the equation
$$\dfrac{5}{7}+2\times\dfrac{18}{7}=k$$
$$k=\dfrac{41}{7}$$
Hence, option 'A' is correct.
Question 2
1 / -0

On the Cartesian plane, Q is the midpoint of the line segment PR.
Find the value of x, y.

A
$$x=3, y=2$$

B
$$x=4, y=2$$

C
$$x=4, y=3\frac{1}{2}$$

D
$$x=8, y=3$$

Solution

by mid point formula
$$x=\dfrac{7+1}{2}=4$$
$$\dfrac{y+4}{2}=3\Rightarrow y=2$$
$$\therefore x=4, y=2$$
Question 3
1 / -0

In the diagram above, KN is a straight line. L and M are two points on KN such that $$KL=LM$$ and $$KM=MN$$. Find the coordinates of L.

A
(3, 2)

B
(3, 4)

C
(4, 4)

D
(4, 2)

Solution

Co-ordinates of M are
$$\left ( \dfrac{0+12}{2},\dfrac{2+10}{2} \right )=(6, 6)$$
Co-ordinates of L are
$$\left ( \dfrac{0+6}{2},\dfrac{2+6}{2} \right )=(3, 4)$$
Question 4
1 / -0

The coordinates of three consecutive vertices of a parallelogram are $$\displaystyle (1, 3), (-1, 2)$$ and $$\displaystyle (2, 5).$$ The coordinates of the fourth vertex are

A
$$\displaystyle (6,4)$$

B
$$\displaystyle (4,6)$$

C
$$\displaystyle (-2,0)$$

D
none of these

Solution

Let the parallelogram be $$ABCD$$ and $$O$$ is the midpoint of both diagonals $$AC$$ and $$DB$$
So, $$AO=OC$$ and $$DO=OB$$.
Let $$A(1,3),B(-1,2),C(2,5)$$ and $$D(x,y)$$
As $$AO=OC$$, $$O$$ is mid point of $$AC$$,
$$h=\cfrac { 1+2 }{ 2 } ,k=\cfrac { 3+5 }{ 2 } $$
$$\Rightarrow h=3/2,\,k=4$$
Hence, coordinate of $$O$$ is $$(3/2,4)$$.
Now, $$BO=OD$$
so by midpoint formula i.e,
$$3/2 = (x+(-1))/2$$ and $$4 = (y+2)/2$$ from these equations we get calculate x and y
So, the coordinates of $$D$$ are $$(4,6)$$.
Question 5
1 / -0

The co-ordinates of the points which divide the line segment joining the points $$(-2, 2)$$ and $$(6, -6)$$ in four equal parts is

A
$$(0, 0), (3, -3), (4, -4)$$

B
$$(0, 0), (2, -2), (4, -4)$$

C
$$(0, 0), (2, -2), (3, -3)$$

D
$$(-1, 1), (2, -2), (4, -4)$$

Solution

Let $$AB$$ be a line segment joining the points $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( -2,2 \right)$$ and $$B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 6,-6 \right)$$ which is divided into $$4$$ equal parts at $$P, Q$$ and $$R$$.
Then $$Q$$ will be the mid point of $$AB$$
We shall apply the section formula for mid point as $$\left( x,y \right) =\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right)$$
i.e. the co-ordinates of $$Q,$$ by mid point formula, is
$$Q\left( \dfrac { -2+6 }{ 2 }, \dfrac { 2-6 }{ 2 } \right) =\left( 2,-2 \right)$$
Again $$P$$ will be the mid point of $$AQ$$
i.e. the co-ordinates of $$P,$$ by mid point formula, is
$$P\left( \dfrac { -2+2 }{ 2 }, \dfrac { 2-2 }{ 2 } \right) =\left( 0, 0 \right)$$
Also $$R$$ will be the mid point of $$BQ$$
i.e, the co-ordinates of $$R,$$ by mid point formula, is
$$R\left( \dfrac { 2+6 }{ 2 }, \dfrac { -2-6 }{ 2 } \right) =\left( 4,-4 \right)$$
$$\therefore$$ The co-ordinates of $$P,Q$$ and $$R$$ are $$(0,0), (2,-2)$$ and $$(4,-4) $$
Question 6
1 / -0

In the diagram, M is the midpoint of PQ
Find the value of k.

A
1

B
2

C
3

D
5

Solution

Using mid-point formulae:
$$\dfrac{1+k}{2}=2$$
$$k=3$$
Question 7
1 / -0

The co-ordinates of the point A which divide the line segment $$GH$$ joining the points $$G(-4,12)$$ and $$H(11,-3)$$ internally in the ratio $$m: n=2:3$$ are

A
$$ A\left( x,y \right) =(1,3)$$

B
$$ A\left( x,y \right) =(2,3) $$

C
$$ A\left( x,y \right) =(2,6) $$

D
$$ A\left( x,y \right) =(1,6)$$

Solution

Given that the point $$A\left( x,y \right)$$ divides the line segment $$GH$$ joining the points
$$G\left( { x }_{ 1 },{ y }_{ 1 } \right) =(-4,12)$$ & $$H\left( { x }_{ 2 },{ y }_{ 2 } \right) =(11,-3)$$ in the ratio $$m:n=2:3$$.
We know that if a point $$A\left( x,y \right)$$ divides the line segment $$GH$$ joining the points
$$G\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ & $$H\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ in the ratio $$m:n$$ then, by the section formula,
$$ x=\dfrac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n }$$ & $$y=\dfrac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n }$$
$$\therefore$$ The co-ordinates of A are
$$ x=\dfrac { 2\times \left( 11 \right) +3\times (-4) }{ 2+3 } =\dfrac { 10 }{ 5 }=2$$
And $$y=\dfrac { 2\times \left( -3 \right) +3\times 12 }{ 2+3 } =\dfrac { 30 }{ 5 } =6$$
$$ \therefore A\left( x,y \right) =(2,6)$$
Question 8
1 / -0

The distance of the point $$Z (-2.4, -1)$$ from the origin is

A
$$2.6\ units$$

B
$$ 4\sqrt { 2.6 }\ units$$

C
$$-2.6\ units$$

D
$$ -4\sqrt { 2.6 }\ units$$

Solution

Given that:
The coordinates of a point Z are $$(-2.4,-1)$$.

To find out:
The distance $$OZ$$, when $$O$$ is the origin.

We know that the distance of a point $$Z(x,y$$) from the origin $$O$$ is $${ d }_{ oz }=\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$$
$$\therefore OZ=\sqrt { { \left( -2.4 \right) }^{ 2 }+{ \left( -1 \right) }^{ 2 } }$$
$$=\sqrt{5.76+1}$$
$$=\sqrt{6.76}$$
$$=2.6\ units$$

Hence, the distance of the point $$Z(-2.4,-1)$$ from the origin is $$2.6\ units$$.
Question 9
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Coordinates of $$P$$ and $$Q$$ are $$(4,-3)$$ and $$(-1, 7)$$. The abscissa of a point $$R$$ on the line segment $$PQ$$ such that $$\dfrac{PR}{PQ}=\dfrac{3}{5}$$ is

A
$$\displaystyle\frac{18}{5}$$

B
$$\displaystyle\frac{17}{5}$$

C
$$1$$

D
$$\displaystyle\frac{17}{8}$$

Solution

Given $$\dfrac{PR}{PQ}=\dfrac{3}{5}$$

$$\Rightarrow \dfrac{PQ}{PR}=\dfrac{5}{3}$$......(1)

Here $$PQ = PR + QR$$
Substituting the value of $$PQ$$ in (1) and subtracting 1 from both sides

$$\dfrac{PR+QR}{PR}-1=\dfrac{5}{3}-1$$

$$\therefore \dfrac{QR}{PR}=\dfrac{2}{3}$$

So the ratio at which $$R$$ divides $$PQ$$ is $$3:2$$

We know that if a point $$R\left( x,y \right)$$ divides the line segment $$PQ$$ joining the points $$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ and $$Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ in the ratio $$m:n$$ then by the section formula.

$$ x=\dfrac { m{ x }_{ 2 } +nx_1 }{ m+n } \quad \& \quad y=\dfrac {m{ y }_{ 2 } +n{y_1}}{ m+n } \quad $$
As we are given the coordinates of $$P$$ and $$Q$$ and also $$\dfrac{QR}{PR}$$,
$$x= \dfrac{3\times (-1)+2\times 4}{3+2}, y=\dfrac{3\times 7+2\times(- 3)}{3+2}$$
$$\therefore x= 1, y=3$$

So the abscissa of point R is 1
Question 10
1 / -0

The co-ordinates of the mid point of segment $$KR$$, where $$K(2.5, -4.3)$$ and $$R(-1.5, 2.7)$$, are

A
$$(0.5, 0.8)$$

B
$$(-0.5, -0.8)$$

C
$$(-0.5, 0.8)$$

D
$$(0.5, -0.8)$$

Solution

Given that $$KR$$ is a line segment joining the points $$K=(2.5, -4.3)$$ and $$R=(-1.5, 2.7)$$
We know that the co-ordinate of the midpoint $$M$$ of a line joining the points $$(x_1, y_1)$$ and $$(x_2,y_2)$$ is given by
$$M=\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 }, \dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right)$$
Let $$M_{KR}$$ be the midpoint of the line segment $$KR$$

$${ \therefore \quad M }_{ KR }=\left( \dfrac {2.5+(-1.5)}{2}, \dfrac{-4.3+2.7}{2} \right) =(0.5,-0.8)$$
Hence, option D is correct.